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Let's have the following problem:

$$U(\boldsymbol{x}) = \left(x_1^\rho + x_2^\rho \right)^{\frac{1}{\rho}} \qquad s.t. \qquad P_1 x_1 + P_2 x_2 \leq M$$

Is it optimal to solve this problem via pure Lagrangian:

$$\mathscr{L} = \left(x_1^\rho + x_2^\rho \right)^{\frac{1}{\rho}} + \lambda \left( M - P_1 x_1 - P_2 x_2 \right) $$

or Lagrangian with non-negativity constraints (or KKT):

$$\mathscr{L} = \left(x_1^\rho + x_2^\rho \right)^{\frac{1}{\rho}} + \lambda \left( M - P_1 x_1 - P_2 x_2 \right) + \mu_1 x_1 + \mu_2 x_2$$

The thing is that analyzing perfect substitutes requires you to use non-negativity constraints to obtain the correct solution.

However, for instance, what if $\rho = 0.9$ instead of $\rho = 1$? Can we say that the optimal solution would always be interior such that we do not need to consider the non-negativity constraints?

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    $\begingroup$ In principle, to mathematical rigor, the condition $\geq0 $ should be set from the beginning, because powers with rational exponent in $\mathbb{R}$ are not defined for negative numbers (only If we have an integer odd number the root is defined also for negative numbers). Otherwise, we can go to complex numbers and complex functions. But, as tdm says, this non negativity constraints are redundant in your CES case, except for $\rho=1$ $\endgroup$ Commented Feb 20 at 16:32

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If $\rho < 1$ then the marginal utility (say with respect to $x_1$ equals): $$ \frac{\partial U}{\partial x_1}(x_1, x_2) = x_1^{\rho - 1} (x_1^\rho + x_2^\rho)^{\frac{1}{\rho}- 1},$$ This tends to $+\infty$ as $x_1$ tends to zero.

As such, it will never be optimal to set $x_1$ equal to zero.

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  • $\begingroup$ I do not understand this explanation fully. I get that big marginal utility incentivizes me to purchase another product stronger than small marginal utility. To be clear, on new hypothetical example, if marginal utility increased to $+ \infty$ as $x_1$ went, for instance, to $-2$, this situation would allow for setting $x_1 = 0$? $\endgroup$
    – Athaeneus
    Commented Feb 20 at 14:02
  • $\begingroup$ The intuition is that setting $x_1 = 0$ is never optimal because increasing $x_1$ a tiny bit (with a corresponding marginal decrease in $x_2$) would lead to an infinite increase in utility. In principle, you can always use the KT conditions and then verify that the constraints $x_1 \ge 0$ and $x_2 \ge 0$ are never binding. $\endgroup$
    – tdm
    Commented Feb 20 at 15:01

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